Question

Solve : ( 1 + tan²θ )( 1 - sin²θ )( 1 + sin²θ )

Answer 1

Given :-

• ( 1 + tan²θ )( 1 - sin²θ )( 1 + sin²θ )

To Do :-

• Solve it.

Solution :-

❍ To calculate the given expression we must know that ::

• 1 + tan²θ  = sec²θ

• ( a + b )( a - b ) = a² - b²

• 1 - sin²θ  = cos²θ

Finding the solution :-

⇝ ( 1 + tan²θ )( 1 - sin²θ )( 1 + sin²θ )

• According to the information we have, we can write the first bracket as sec²θ  

⇝ sec²θ( 1 - sin²θ )( 1 + sin²θ )  

• According to the property of ( a + b )( a - b ) = a² - b²  

⇝ sec²θ( 1² - sin²θ )

⇝ sec²θ × ( 1 - sin²θ )

• According to the information we have, we can write the bracket as cos²θ  

⇝ sec²θ × cos²θ

⇝ 1/cos²θ × cos²θ

⇝ 1

More to know :-

$\boxed{\begin{minipage}{7 cm}Fundamental Trigonometric Identities \\ \\ \sin^2\theta + \cos^2\theta=1 \\ \\1+\tan^2\theta = \sec^2\theta \\ \\1+\cot^2\theta = \text{cosec}^2 \, \theta \end{minipage}}$

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Answer 2

Answer:

Given :-

( 1 + tan²θ )( 1 - sin²θ )( 1 + sin²θ )

To Do :-

Solve it.

Solution :-

❍ To calculate the given expression we must know that ::

1 + tan²θ  = sec²θ

( a + b )( a - b ) = a² - b²

1 - sin²θ  = cos²θ

Finding the solution :-

⇝ ( 1 + tan²θ )( 1 - sin²θ )( 1 + sin²θ )

According to the information we have, we can write the first bracket as sec²θ  

⇝ sec²θ( 1 - sin²θ )( 1 + sin²θ )  

According to the property of ( a + b )( a - b ) = a² - b²  

⇝ sec²θ( 1² - sin²θ )

⇝ sec²θ × ( 1 - sin²θ )

According to the information we have, we can write the bracket as cos²θ  

⇝ sec²θ × cos²θ

⇝ 1/cos²θ × cos²θ

⇝ 1